Experiment 20 impulse and momentum relationship

Impulse and momentum dodgeball example (video) | Khan Academy

experiment 20 impulse and momentum relationship

during the collision and to compare these values as a test of the impulse- momentum theorem. <. To compare the action and reaction forces that two carts exert. Read Watch Interact · Practice Review Test . The equation is known as the impulse-momentum change equation. The law can be The impulse experienced by the object equals the change in momentum of the object. In equation . Airbags are used so that the momentum of the person transfers into the this experiment does not support the impulse-momentum theorem.

Our face exerted a force on the ball to the right. That's why the impulse on the ball is to the right. The impulse on this person's face is to the left, but the impulse on the ball is to the right, because the ball was initially going left and it had a force on it to the right that made it recoil and bounce back to the right.

That's why this impulse has a positive direction to it. Now, if you've been paying attention, you might be like, wait a minute, hold on. What we really did was we found the change in momentum of the ball, and when we do that, what we're finding is the net impulse on the ball. In other words, the impulse from all forces on the ball.

But what this question was asking for was the impulse from a single force. The impulse from just the person's face. Now, aren't there other forces on this ball? Isn't there a force of gravity? And if there is, doesn't that mean what we really found here wasn't the impulse from just our face, but the impulse from the person's face and the force of gravity during this time period?

And the answer is no, not really, for a few reasons.

Momentum and Impulse Connection

Most important reason being that, what I gave you up here was the initial horizontal velocity. This 10 meters per second was in the X direction, and this five meters per second, I'm assuming is also in the X direction.

When I do that, I'm finding the net impulse in the X direction, and there was only one X directed force during this time and that was our face on the ball, pushing it to the right. There was a force of gravity. That force of gravity was downward. But what that force of gravity does, it doesn't add or subtract any impulse in the X direction.

It tries to add impulse in the downward direction, in the Y direction, so it tries to add vertical component of velocity downward, and so we're not even considering that over here. We're just gonna consider that we're lookin' at the horizontal components of velocity.

Impulse and momentum dodgeball example

How much velocity does it add vertically, gravity? Typically, not much during the situation, because the time period during which this collision acted is very small and the weight of this ball, compared to the force that our face is acting on the ball with, the weight is typically much smaller than this collision force.

So that's why, in these collision problems, we typically ignore the force of gravity. So, we don't have to worry about that here. That's not actually posing much of a problem. We did find the net impulse in the X direction since our face was the only X directed force, this had to be the impulse our face exerted on the ball. Now, let's solve one more problem. Let's say we wanted to know: What was the average force on this person's face from the ball?

Well, we know the net impulse on the ball, that means we can figure out the net force on the ball, because I can use this relationship now. Since I know that the net impulse on the ball in the X direction should just equal the net force on the ball in the X direction, multiplied by the time interval during which the force was applied, I can say that the net impulse on the ball was three kilogram meters per second, and that should equal the net force on the ball in the X direction, which was supplied by, unfortunately, this person's face, multiplied by the time interval, which is 0.

So, now I can solve. The force, the net force on the ball, during this time interval in the X direction, was three, divided by. If I take three kilogram meters per second and I divide by. We got a positive number, and that makes sense, because this person's face exerted a positive force on this ball, 'cause the force was exerted to the right.

So, these were positive, and the impulse from the face on the ball should be going the same direction as the force from the face on the ball. So, this is the force on the ball by the person's face, but notice this question is asking: What was the average force on the person's face from the ball?

Not on the ball by the face. You might think, oh no, we gotta start all over, we solved for the wrong question, but we're in luck. Newton's third law says that the force on the face from the ball should be equal and opposite. So, this force on the face from the ball has got to be equal and opposite to the force on the ball from the face, so that's what we found here, the force on the ball from the face, that means the force on the face from the ball is gonna have the same size.

It's gonna be Newtons. It's just gonna be directed in the leftward direction, that means it's gonna be a negative force, so technically, you could say, this would be negative Newtons on the face from the ball.

So, to recap, the impulse, from an individual force, is defined to be that force, multiplied by the time interval during which that force is applied. And if you're talkin' about the net force in a given direction, multiplied by the time interval, you'd be finding the net impulse in that direction, and this also happens to equal the change in momentum in that direction. So, in other words, if there is a net impulse in a given direction, there's gotta be a change in momentum in that direction by the same amount.

And one convenient way to remember how are these related, is you could use the pneumonic device, Jape Fat. I have no idea what Jape Fat means, but it helped me remember that the net impulse equals the change in momentum, and that also equals the net force, multiplied by the time interval, during which that force was applied.

And finally, remember that during these collisions, there's always an equal and opposite force exerted on the two objects participating in the collision. Choose a string length so that the cart can roll freely with the cord slack for most of the track length, but be stopped by the cord before it reaches the end of the track.

experiment 20 impulse and momentum relationship

When the cord is stretched to maximum extension the cart should not be closer than 0. Clickthento zero the Force Sensor. Do not zero the MD. Practice releasing the cart so it rolls toward the Motion Detector, bounces gently, and returns to your hand.

The Force Sensor must not shift and the cart must stay on the track. Arrange the cord and string so that when they are slack they do not interfere with the cart motion. You may need to guide the string by hand, but be sure that you do not apply any force to the cart or Force Sensor. Keep your hands away from between the cart and the Motion Detector. Click to take data; roll the cart and confirm that the Motion Detector detects the cart throughout its travel.

Inspect the force data.

Momentum and Impulse Connection

If the peak is flattened, then the applied force is too large. Roll the cart with a lower initial speed. If the velocity graph has a flat area when it crosses the x-axis, the Motion Detector was too close and the run should be repeated.

experiment 20 impulse and momentum relationship

Once you have made a run with good distance, velocity, and force graphs, analyze your data. To test the impulse-momentum theorem, you need the velocity before and after the impulse.

Choose a time interval just before the bounce when the speed was approximately constant, and drag the mouse pointer across the interval. Click the Statistics button,and read the average velocity. Record the value in your data table. In the same manner, determine the average velocity just after the bounce and record the value in your data table. Remove the floating boxes by clicking the gray close box in their upper right corners.

There are two ways to calculate the impulse.

Momentum and Impulse Lab: PASCO

Calculus tells us that the expression for the impulse is equivalent to the integral [area under the curve] of the force vs. Find the area under the force vs. Record the value of the integral in the impulse column of your data table. On the force vs.

experiment 20 impulse and momentum relationship

Find the average value of the force by clicking the Statistics button. Record the values in your data table. Add a cart mass and perform a second trial by repeating Steps 10 — Record the information in your data table. Change the elastic material attached to the cart. Use a new material, different elastic band, or attach two elastic bands in parallel. Repeat Steps 10 — 13, record the information in your data table. Change in momentum is a measure of what resulted: Include in your lab writeup the relevant plots and data analysis boxes for trial 1.

Impulse - Linear Momentum, Conservation, Inelastic & Elastic Collisions, Force - Physics Problems

Calculate the change in velocities and record in the data table. From the mass of the cart and change in velocity, determine the change in momentum as a result of the impulse. Make this calculation for each trial and enter the values in the second data table. For the average force non-calculus method, determine the impulse for each trial from the average force and time interval values. Record these values in your data table. If the impulse-momentum theorem is correct, the change in momentum will equal the impulse for each trial.

Experimental measurement errors, along with friction and shifting of the track or Force Sensor, will keep the two from being exactly the same. One way to compare two or moe values that are supposed to be equal is to find their percentage difference. How close are your values, percentage-wise? Do your data support the impulse-momentum theorem? Look at the shape of the last force vs.

Is the peak value of the force significantly different from the average force? Is there a way you could deliver the same impulse with a much smaller force?